Why I use function generic parameters as the other type's generic has compile error?












0















I define a function with signature <U extends UU, UU> and a Type type T04<F extends FF, FF> = T03<F> extends T03<FF> ? number : string;,then I use the U and UU as this Type's generic type parameters. As normal pass "aaa" and string to the T04 I will get the number type ,so that if pass U and UU to the T04,I will get the number type too ,but not get my expected type. Why is this happening?



Typescript:3.2.0



function fn1<U extends UU, UU>() {

  type T03<T> = {

    action: T;

  };

type T04<F extends FF, FF> = T03<F> extends T03<FF> ? number : string;



let v2: T04<"aaa", string> = 222;

let v3: T04<U, UU> = 78787;


}



expect v3' type is number






typescript







share|improve this question



























    0















    I define a function with signature <U extends UU, UU> and a Type type T04<F extends FF, FF> = T03<F> extends T03<FF> ? number : string;,then I use the U and UU as this Type's generic type parameters. As normal pass "aaa" and string to the T04 I will get the number type ,so that if pass U and UU to the T04,I will get the number type too ,but not get my expected type. Why is this happening?



    Typescript:3.2.0



    function fn1<U extends UU, UU>() {
    
  type T03<T> = {
    
    action: T;
    
  };
    
type T04<F extends FF, FF> = T03<F> extends T03<FF> ? number : string;
    

    
let v2: T04<"aaa", string> = 222;
    
let v3: T04<U, UU> = 78787;


    }



    expect v3' type is number






    typescript







    share|improve this question

























      0












      0








      0








      I define a function with signature <U extends UU, UU> and a Type type T04<F extends FF, FF> = T03<F> extends T03<FF> ? number : string;,then I use the U and UU as this Type's generic type parameters. As normal pass "aaa" and string to the T04 I will get the number type ,so that if pass U and UU to the T04,I will get the number type too ,but not get my expected type. Why is this happening?



      Typescript:3.2.0



      function fn1<U extends UU, UU>() {
      
  type T03<T> = {
      
    action: T;
      
  };
      
type T04<F extends FF, FF> = T03<F> extends T03<FF> ? number : string;
      

      
let v2: T04<"aaa", string> = 222;
      
let v3: T04<U, UU> = 78787;


      }



      expect v3' type is number






      typescript







      share|improve this question














      I define a function with signature <U extends UU, UU> and a Type type T04<F extends FF, FF> = T03<F> extends T03<FF> ? number : string;,then I use the U and UU as this Type's generic type parameters. As normal pass "aaa" and string to the T04 I will get the number type ,so that if pass U and UU to the T04,I will get the number type too ,but not get my expected type. Why is this happening?



      Typescript:3.2.0



      function fn1<U extends UU, UU>() {
      
  type T03<T> = {
      
    action: T;
      
  };
      
type T04<F extends FF, FF> = T03<F> extends T03<FF> ? number : string;
      

      
let v2: T04<"aaa", string> = 222;
      
let v3: T04<U, UU> = 78787;


      }



      expect v3' type is number






      typescript




      typescript






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 2 at 3:05









      MiloMilo

      12




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